The area of the square: [tex]A_S=x^2[/tex]
The area of the rectangle: [tex]A_R=x(x+6)[/tex]
The area of the rectangle is 4 times the areao f the square:
[tex]A_R=4A_S[/tex]
therefore
[tex]x(x+6)=4x^2\\\\(x)(x)+(x)(6)=4x^2\\\\x^2+6x=4x^2\ \ \ \ |-4x^2\\\\-3x^2+6x=0\\\\-3x(x-6)=0\iff-3x=0\ \vee\ x-6=0\\\\x=0\ \vee\ x=6[/tex]
a) the length of the square x = 6
[tex]x+6=6+6=12[/tex]
b) the width of the rectangle x + 6 = 12
